Question

How many liters of 0.3m h3po4 are needed to neutralize 3.5l of 3m naoh

Answer 1

11.7 L

Step-by-step explanation:

How many liters of 0.3M H₃PO₄ are needed to neutralize 3.5L of 3M NaOH?

Step 1: Write the balanced equation

H₃PO₄ + 3 NaOH = Na₃PO₄ + 3 H₂O

Step 2: Calculate the reacting moles of sodium hydroxide

3.5 L of 3 M NaOH were employed. The reacting moles of NaOH are:

`3.5L * (3mol)/(L) = 10.5 mol`

Step 3: Calculate the reacting moles of phosphoric acid

The molar ratio of H₃PO₄ to NaOH is 1:3. The reacting moles of H₃PO₄ are (1/3) × 10.5 mol = 3.5 mol

Step 4: Calculate the required liters of phosphoric acid

3.5 moles of 0.3 M H₃PO₄ were employed. The required volume of H₃PO₄ is:

`3.5mol * (1L)/(0.3mol) =11.7 L`

Answer 2

`V_(acid)=11.7L`

Step-by-step explanation:

Hello,

In this case, given the reaction:

`3NaOH+H_3PO_4\rightarrow Na_3PO_4+3H_2O`

We notice a 3:1 molar ratio between sodium hydroxide and phosphoric acid, therefore, at the equivalence point we have:

`n_(base)=3*n_(acid)`

That in terms of molarity is:

`M_(base)V_(base)=3*M_(acid)V_(acid)`

Se we solve for the volume of acid:

`V_(acid)=(M_(base)V_(base))/(3*M_(acid)) =(3M*3.5L)/(3*0.3M)\\ \\V_(acid)=11.7L`

Best regards.